Let $A = \mathbb{R}^3 \cap \{(x, \pm{x}, y) : x^2 + y^2 = 1\}$. We have to prove that $A$ (having removed two points) is a differentiable submanifold of dimension $1$.
I see that $A$ is a submanifold of dimension $2$ because : If we note $f(w) = f(x,y) = x^2 + y^2 - 1$, $A$ is not a submanifold if $f'(w) = (2x,2y)=(0,0)$ which is impossible because $x^2 + y^2 = 1$.
But how to evaluate this in dimension $1$ ? Normally, we should have a submanifold only if $A$ having removed two points.
Rename $y \leftarrow z$ to avoid confusion. The set is the intersection of the cylinder $$x^2+z^2=1, y \in R$$ with the planes $$y=x$$ and $$y=-x.$$
Its two rings touching twice on the $z$-axis. In fact, one is obtained by rotating one around the $z$-axis. To avoid the "X" shape at the intersections we remove them. Then we end up with 4 half circles (without end points). Each a 1-dimensional manifold.
This was to illustrate the geometry. An easier way of seeing this is by parameterizing the mentioned half circles. Someone did so in the comment for one piece. By switching signs in the first two coordinates you get the other ones.